Let $a_1,$ $a_2,$ $a_3,$ $\dots$ be a sequence of real numbers satisfying
\[a_n = a_{n - 1} a_{n + 1}\]for all $n \ge 2.$  If $a_1 = 1 + \sqrt{7}$ and $a_{1776} = 13 + \sqrt{7},$ then determine $a_{2009}.$
Answer: From the given recursion,
\[a_{n + 1} = \frac{a_n}{a_{n - 1}}.\]Let $a = a_1$ and $b = a_2.$  Then
\begin{align*}
a_3 &= \frac{a_2}{a_1} = \frac{b}{a}, \\
a_4 &= \frac{a_3}{a_2} = \frac{b/a}{b} = \frac{1}{a}, \\
a_5 &= \frac{a_4}{a_3} = \frac{1/a}{b/a} = \frac{1}{b}, \\
a_6 &= \frac{a_5}{a_4} = \frac{1/b}{1/a} = \frac{a}{b}, \\
a_7 &= \frac{a_6}{a_5} = \frac{a/b}{1/b} = a, \\
a_8 &= \frac{a_7}{a_6} = \frac{a}{a/b} = b.
\end{align*}Since $a_7 = a = a_1$ and $a_8 = b = a_2,$ and each term depends only on the two previous terms, the sequence is periodic from here on.  Furthermore, the length of the period is 6.  Therefore, $a_6 = a_{1776} = 13 + \sqrt{7}$ and $a_{2009} = a_5.$  Also, $a_7 = a_1,$ and
\[a_7 = \frac{a_6}{a_5}.\]Hence,
\[a_5 = \frac{a_6}{a_7} = \frac{13 + \sqrt{7}}{1 + \sqrt{7}} = \frac{(13 + \sqrt{7})(\sqrt{7} - 1)}{(1 + \sqrt{7})(\sqrt{7} - 1)} = \frac{-6 + 12 \sqrt{7}}{6} = \boxed{-1 + 2 \sqrt{7}}.\]